Recall that \(G\sim \det (G)^{1/3} D(G)\) , where \(D(G)\in \lbrace \mathrm {SO}_3(C),\mathrm {SL}_3(C) \rbrace \) . By Theorem REF , \(D(G)=\mathrm {SO}_3(C)\)  if and only if
there exist \(T\in \mathrm {GL}_3(\mathbf {k})\) , \(\lambda \in \mathbf {k}^*\)  (with \(\lambda ^3\in \det (G)(\mathbf {k})\) ) and \(B\in \mathrm {SO}_3(\mathbf {k})\)  such that (REF ) holds. So the first two items are equivalent. If the second item holds, then the third item holds. Conversely assume that the third item holds.
Consider \(\overline{G}\) , the Galois group over \(\overline{\mathbf {k}}\) . By Theorem REF , \(\overline{G}\sim \det (\overline{G})^{1/3} \mathrm {SO}_3(C)\) .
By Galois correspondence, see [1]}, \(\overline{G}\)  is a subgroup of \(G\)  and \(G/\overline{G}\)  is finite. Since \(G\sim \det (G)^{1/3} D(G)\) , where \(D(G)\in \lbrace \mathrm {SO}_3(C),\mathrm {SL}_3(C) \rbrace \) , this shows that \(D(G)\)  cannot be \(\mathrm {SL}_3(C)\) . Then \(D(G)= \mathrm {SO}_3(C)\) .
